7-56. Trigonometric substitutions Evaluate the following integ...

Question

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

36. ∫[8√2 to 16] 1/√(x² - 64) dx

Accepted Answer

Welcome back, everyone. In this problem, evaluate the integral using trigonometric substitution of 1 divided by the square root of X2 minus 100 with respect to X on the interval from 10 root 2 to 20. Now, for us to evaluate this integral, let's first start by doing a few substitutions, OK? So let's let I be equal to our integral, and that is going to make it easier to refer to as we go along in solving our problem, OK. And not to involves some trigonometric terms. Let's let x be equal to 10 multiplied by the C of theta. In that case, The derivative of X with respect to theta is going to be equal to 10 sec theta. And theta, OK, which means that DX is going to be equal to 0 theta theta, D theta. No, we've been able to write DX in terms of theta and X in terms of theta, so we have all we need for our integra. No, we only need to change the bones and then substitute. Because changing our bones. OK, these original bones are the bones for eggs. Then for X. Eals 10 root 2, OK. Since we know X equals 10 theta, then that means 10 sec theta. OK. Would be equal to 10 root 2. If we can 1 from both sides, OK, and then here, so, that means sick theater. Would be equal to root 26 theta equals root 2, and if 6 theta equals root 2, then theta is going to be equal to a 4th of pi. On the other hand, for X equals 20, OK. X equals 20, then that means 10 theater will be equal to 20, which means sex theater would be equal to 2 if we divide both sides by 10. And if 6 theta equals 2, then that means theater. Is going to be equal to 1/3 of pie. So now we also have the bones or the bones that we need for the trigonometric substitution. So by substituting into our integral. OK, so no, by substituting. Then we're going to get I. To be equal to the integral, OK, between 4th of pi and a 1/3 of pi. Of one divided by the square root, OK. Of 10 sec theta squared minus 100 multiplied by 10 sec theta 10 theta with respect to theta. Now where do we go from here? Well, let's do a little work, OK? So now, notice that. Here, if we expand the term under our square root. Then we're gonna have it to become. 106th grade theater. OK, -100. OK. And if we expand that a little further, let me see if I can write this to the right, OK. Then we could rewrite it, OK. As 100, we could factor 100, so this is gonna be the square root of 100 multiplied by 6 square theta minus 1. OK? Now, we're still multiplying that by 10, 6, theta 10 theta with respect to theta. Now, why is this helpful? Well, we know that the square root of 100 is 10 and 6 square theater minus 1 as an identity, OK, let's put that here, is equal to 1qua theater. Times great theater equals 6 qua theta minus 1. So in that case, we could substitute that into our expression and when we do that, OK, we're now we're going to end up with the integral between 3, 4th of pi and 3 of pi of 1 divided by the square root of 100, 10 multiplied by the square root of 10 square the theta, OK, multiplied by 10 sec theta 10 theta with respect to theta. No, why is this helpful? Well, notice that we can cancel 10 and we can cancel 1010 theater. Sorry, no, that's gonna be giving us the integral between 1/3 of pie and a 14th of pie of sec theater with respect to theater. Now we can evaluate the integral. Because no, this means that I, when we integrate sec theater with respect to theta is going to be equal to the natural log of sec theta plus 10 theta, OK, between the bones of 1/4 of pi and 13 of pi, OK? Now, when we apply our bones here. That means we're going to get the natural log. Of a of a third of pi, OK, plus the tangent of 1/3 of pi, OK, and that's the absolute value of those terms minus the natural log of the absolute value of the 24th of pi plus the tangent of 4th of pi, OK. And now when we go ahead and expand that, OK, we can solve for or see can terms. So that's gonna be the natural log of 2 + route 3, OK, minus the natural log of route 2 + 1. Now remember by the laws of logs, if we're subtracting two logs with the same base, then we could rewrite the the the terms as a quotient. So that's the natural log of 2 + route 3 divided by route 2 + 1. And now since the argument is positive, that means the absolute value sign is unnecessary. So then the integral is going to be equal to the natural log of 2 + route 3 divided by route 2 + 1. This is our integral using trigonometric substitution. Thanks for watching everyone. I hope this video helped.

7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.36. ∫[8√2 to 16] 1/√(x² - 64) dx

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7-56. Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.
36. ∫[8√2 to 16] 1/√(x² - 64) dx